📝 How to evaluate surface integrals (16 MCQs)

📖 From Calculus • 16. Topics in vector Calculus • 16 questions available

←Previous Topic:Surface integral definition📝 How to evaluate surface integralsSurface integrals over different surfaces z = g(x,y), y = g(x,z), and x = g(y,z):Next Topic→

What is How to evaluate surface integrals?

How to evaluate surface integrals:
Parameterize surface, compute $\mathbf{r}_u \times \mathbf{r}_v$ , find its magnitude, substitute into integrand, and integrate over the parameter domain.

Example:
For $f = x^2$ over the cylinder $x^2 + y^2 = 1$ , $0 \le z \le 1$ , use $\mathbf{r}(\theta,z) = \langle \cos\theta, \sin\theta, z \rangle$ , $|\mathbf{r}_\theta \times \mathbf{r}_z| = 1$ , integral = $\int_0^{2\pi} \int_0^1 \cos^2\theta \, dz d\theta = \pi$ .

Reason:
Step-by-step evaluation, like line integrals, builds computational skills for physical applications.

Easy

Medium

Hard

📝 All How to evaluate surface integrals MCQs

Q1. A surface $S$ is given by $z=4-x^2-y^2$ above the disk $x^2+y^2\leq1$ . Which setup correctly represents the scalar surface integral $\iint_S z\,dS$ after projection onto the $xy$ -plane?

\int_0^{2\pi}\int_0^1 (4-r^2)r\,dr\,d\theta

\int_0^{2\pi}\int_0^1 (4-r^2)\sqrt{1+4r^2}\,r\,dr\,d\theta

✅

\int_0^{2\pi}\int_0^1 (4-r^2)\sqrt{1+r^2}\,r\,dr\,d\theta

\int_0^{2\pi}\int_0^1 4r\sqrt{1+4r^2}\,dr\,d\theta

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For a graph $z=f(x,y)$ , the surface element is $dS=\sqrt{1+f_x^2+f_y^2}\,dA$ . Here $f_x=-2x$ and $f_y=-2y$ , so the factor becomes $\sqrt{1+4x^2+4y^2}=\sqrt{1+4r^2}$ . The integrand is $z=4-r^2$ , and $dA=r\,dr\,d\theta$ .

Q2. For a parametrized surface $\mathbf r(u,v)$ , which expression represents the scalar surface element used to evaluate $\iint_S g\,dS$ ?

\left|\mathbf r_u\times\mathbf r_v\right|\,du\,dv

✅

\mathbf r_u\cdot\mathbf r_v\,du\,dv

\left|\mathbf r_u+\mathbf r_v\right|\,du\,dv

\left|\mathbf r_u\times\mathbf r_v\right|^2\,du\,dv

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The magnitude of the cross product $\mathbf r_u\times\mathbf r_v$ gives the area-scaling factor from the parameter domain to the surface. Therefore $dS=\left|\mathbf r_u\times\mathbf r_v\right|du\,dv$ . The dot product instead measures the angle-related interaction of tangent vectors and does not generally give the required surface-area element.

Q3. A student evaluates $\iint_S f(x,y,z)\,dS$ over a surface described by $z=g(x,y)$ but replaces $dS$ with $dx\,dy$ . What is the main conceptual error?

A.The student forgot to substitute

z=g(x,y)

into

f

B.The student ignored the stretching of the horizontal area element onto the tilted surface. ✅

C.The student used a scalar integral instead of a vector integral.

D.The student must always use polar coordinates for a surface integral.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The projection $dx\,dy$ measures area in the horizontal plane, not directly on the surface. A tilted or curved surface generally has greater area than its projection. The factor $\sqrt{1+g_x^2+g_y^2}$ accounts for this geometric stretching and must multiply the projected area element.

Q4. Consider the plane $z=2x+3y+1$ over a region $D$ in the $xy$ -plane. Which observation allows the surface integral $\iint_S 1\,dS$ to be evaluated most efficiently?

A.The surface-area factor is constant because the plane has constant partial derivatives. ✅

B.The surface-area factor is zero because the plane has no curvature.

C.The surface-area factor equals

1+2+3

D.The surface-area factor depends on

x

and

y

because every plane has variable slope.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For $z=2x+3y+1$ , the derivatives are $z_x=2$ and $z_y=3$ , both constants. Thus $dS=\sqrt{1+2^2+3^2}\,dA=\sqrt{14}\,dA$ . The entire surface area is therefore $\sqrt{14}$ times the area of its projection, making the evaluation especially direct.

Q5. A hemispherical surface is described by $x^2+y^2+z^2=a^2$ , $z\geq0$ . If the density depends only on height through $\rho=z$ , which strategy best simplifies $\iint_S \rho\,dS$ ?

A.Convert the entire sphere to Cartesian coordinates immediately.

B.Use symmetry and spherical coordinates so that

z=a\cos\phi

and

dS=a^2\sin\phi\,d\phi\,d\theta

. ✅

C.Replace

dS

dz\,d\theta

D.Treat

z

as constant because the surface is spherical.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Spherical coordinates align naturally with the geometry. On the hemisphere, $z=a\cos\phi$ , while the spherical surface element is $dS=a^2\sin\phi\,d\phi\,d\theta$ . This converts the surface integral into a simple angular integral and avoids the complicated square-root factor produced by a Cartesian graph representation.

Q6. A metal sheet occupies the portion of the plane $z=x+y$ above the square $0\leq x\leq2,\ 0\leq y\leq2$ . Its surface density is proportional to $x+y$ . Which expression correctly models its total mass?

\iint_D (x+y)\,dA

\iint_D (x+y)\sqrt{3}\,dA

✅

\iint_D (x+y)\sqrt{1+x^2+y^2}\,dA

\iint_D (x+y)(1+1)\,dA

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The mass is the density integrated with respect to actual surface area. Since $z=x+y$ , we have $z_x=1$ and $z_y=1$ , giving $dS=\sqrt{1+1^2+1^2}\,dA=\sqrt3\,dA$ . Therefore the density $x+y$ must be multiplied by $\sqrt3$ before integrating over the square.

Q7. For $S:z=\sqrt{9-x^2-y^2}$ , the upper hemisphere of radius $3$ , a student argues that using $z=f(x,y)$ must produce an infinite surface-area factor near the boundary because $f_x$ and $f_y$ become large. Which conclusion is most accurate?

A.The argument is correct, so the surface area is infinite.

B.The graph method fails completely for hemispheres.

C.The derivatives become large, but the resulting surface element is integrable and gives the finite hemisphere area. ✅

D.The surface element becomes zero near the boundary and cancels the divergence.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The derivatives of $f=\sqrt{9-x^2-y^2}$ do become unbounded as $x^2+y^2\to9$ , but that does not imply that the total surface area diverges. In polar coordinates the surface element simplifies to $3r/\sqrt{9-r^2}\,dr\,d\theta$ , whose integral from $0$ to $3$ is finite.

Q8. A student evaluates the surface $z=x^2+y^2$ over the unit disk and writes $dS=\sqrt{1+2x+2y}\,dA$ . What mistake has been made?

A.The derivatives should be squared before being added. ✅

B.The function must first be converted to spherical coordinates.

C.The surface element for every paraboloid is simply

dA

D.The student should use

1+x^2+y^2

instead.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For $z=x^2+y^2$ , the partial derivatives are $z_x=2x$ and $z_y=2y$ . The correct factor is therefore $\sqrt{1+(2x)^2+(2y)^2}=\sqrt{1+4x^2+4y^2}$ . The error comes from adding the derivatives directly rather than squaring them as required by the surface-area formula.

Q9. A graph of a surface shows a nearly horizontal patch in one region and a steeply tilted patch of equal projected area in another. If the surface density is constant, which region contributes more to the surface integral?

A.The nearly horizontal region, because its projection is larger.

B.The steep region, because the same projected area corresponds to greater actual surface area. ✅

C.Both contribute equally because their projected areas are equal.

D.Neither can contribute unless the surface is closed.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Equal projected areas do not imply equal actual surface areas. A steeply tilted patch is stretched relative to its projection, so its surface-area factor $\sqrt{1+f_x^2+f_y^2}$ is larger. With constant density, the contribution is proportional to actual area, making the steeper region contribute more.

Q10. A contour-style graph indicates that $z=f(x,y)$ is rotationally symmetric about the $z$ -axis, and the density also depends only on $x^2+y^2$ . Which coordinate choice is most likely to reduce the evaluation effort?

A.Cartesian coordinates, because symmetry should be avoided.

B.Polar coordinates in the

xy

-plane, because both the surface and density are rotationally symmetric. ✅

C.Only spherical coordinates, regardless of the surface equation.

D.A single

z

-integration, because the surface is two-dimensional.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Rotational symmetry about the $z$ -axis strongly suggests polar coordinates. Writing $x=r\cos\theta$ and $y=r\sin\theta$ converts $x^2+y^2$ to $r^2$ , while the angular dependence often disappears. This can reduce a two-dimensional surface integral to a simpler radial integral multiplied by the angular interval.

Q11. A hemispherical tank has surface density proportional to $z^2$ . An engineer can evaluate the mass either by projecting onto the $xy$ -plane or by using spherical coordinates. Which comparison is correct?

A.Both methods must give the same value when their parameterizations cover exactly the same surface. ✅

B.The projection method always gives a larger value because it ignores curvature.

C.Spherical coordinates necessarily give a smaller value because

z\leq a

D.The two methods cannot agree because their surface elements are different.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Different coordinate systems or parameterizations describe the same geometric surface, so a correctly constructed integral must produce the same physical mass. The formulas may look very different because their surface elements are expressed differently, but both account for the same actual area and density.

Q12. A surface patch is parametrized by $\mathbf r(u,v)=(u,v,u^2+v^2)$ for $0\leq u,v\leq1$ . A student computes $\mathbf r_u\times\mathbf r_v=(-2u,-2v,1)$ but uses its first component alone when forming $dS$ . Why is this invalid?

A.Only the first component determines surface area.

B.Surface area requires the magnitude of the full cross product, which accounts for both tangent directions and their relative orientation. ✅

C.The cross product should be replaced by the dot product.

D.The first component is sufficient only when

u=v

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The cross product vector is normal to the surface, and its magnitude gives the area of the parallelogram generated by the two tangent vectors. Therefore $dS=\sqrt{(-2u)^2+(-2v)^2+1^2}\,du\,dv=\sqrt{1+4u^2+4v^2}\,du\,dv$ . Using one component discards essential geometric information.

Q13. Suppose $S$ is the portion of $z=5-x^2-y^2$ above the disk $x^2+y^2\leq4$ , and the density is $\rho=z$ . Which sequence of steps is mathematically sound?

A.Use polar coordinates, substitute

z=5-r^2

, compute

dS=\sqrt{1+4r^2}\,r\,dr\,d\theta

, then integrate over

0\leq r\leq2

. ✅

B.Use

dS=r\,dr\,d\theta

, because polar coordinates automatically include surface curvature.

C.Integrate

z

from

0

5-r^2

, treating the surface as a volume.

D.Use

dS=\sqrt{1+2r}\,r\,dr\,d\theta

, because the derivative of

r^2

2r

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The correct procedure separates the geometry from the density. The graph has derivatives $-2x$ and $-2y$ , producing the factor $\sqrt{1+4r^2}$ . Polar coordinates contribute $r\,dr\,d\theta$ , while the density becomes $5-r^2$ . The disk radius determines $0\le r\le2$ .

Q14. Two methods are proposed for evaluating the area of a planar surface patch: Method I projects it onto the $xy$ -plane and uses a constant factor $C$ ; Method II directly parametrizes the plane and computes $|\mathbf r_u\times\mathbf r_v|$ . If both methods describe the same patch correctly, what must be true?

A.The two answers may differ because projection changes the surface.

B.The answers must agree, and

C

represents the constant stretching factor between the plane and its projection. ✅

C.Method II must always produce twice the result of Method I.

D.The projection method is valid only for horizontal planes.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For a plane, the slopes are constant, so the ratio between actual surface area and projected area is constant. Thus Method I uses a constant stretching factor. A correct parametrization in Method II must encode exactly the same geometric stretching through the cross-product magnitude, so both approaches necessarily yield the same area.

Q15. A surface is the graph $z=f(x,y)$ over a region $D$ , and $f_x^2+f_y^2$ is constant and equal to $8$ . If the surface density is also constant $k$ , what is the most efficient expression for its total mass?

k\operatorname{Area}(D)

3k\operatorname{Area}(D)

k\sqrt8\operatorname{Area}(D)

k\sqrt9\operatorname{Area}(D)

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Since $f_x^2+f_y^2=8$ , the surface element is $dS=\sqrt{1+8}\,dA=3\,dA$ . With constant density $k$ , the mass becomes $k\iint_S dS=3k\operatorname{Area}(D)$ . This illustrates that constant gradient magnitude allows the surface integral to collapse to a simple area calculation.

Q16. A surface integral over a closed spherical surface has integrand $x+y+z$ , and the density is otherwise uniform. Without carrying out a lengthy parametrization, what can be concluded about the integral?

A.It equals the sphere's surface area.

B.It is positive because

x+y+z

is positive on the upper hemisphere.

C.It is zero because contributions from opposite points cancel. ✅

D.It equals three times the sphere's surface area.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Every point $(x,y,z)$ on a sphere has an antipodal point $(-x,-y,-z)$ with the same surface-area contribution but opposite value of $x+y+z$ . Pairing all such points causes exact cancellation. Therefore the integral is zero, even though the integrand is not identically zero.

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